Calculating Maximum Dive Depth: Tips for Cliff Divers | Get the Facts Here!

[Charles123]

How can one calculate the maximum depth a diver (jumping to water, like cliff diving) reaches if he stays in the same position that he used when entering the water, and does nothing to stop more quickly?
Thank you
Regards

 

[Simon Bridge]

Back-of-envelope ... the diver has an initial velocity and a bouyancy: use kinematics.
Assumes negligible drag and no losses entering the water.

Fuid models can get as complicated as you need them. Did you have a particular situation in mind?

 

[russ_watters]

When I was a kid, I was negatively buoyant...

 

[sophiecentaur]

Once a diver has reached a particular depth, he/she will become negatively bouyant as the air in the lungs is compressed. For most people, this depth will only be two or three metres, max. After that depth, they would just keep sinking (a very handy fact for free divers).
The question would apply for an object which was incompressible and streamlined. In that case you could equate the Kinetic Energy with FD where F is the bouyancy force and D is the depth reached. But there would always be some energy losses so the depth would be less than this calculation would suggest.

 

[Borek]

Once a diver has reached a particular depth, he/she will become negatively bouyant as the air in the lungs is compressed. For most people, this depth will only be two or three metres, max. After that depth, they would just keep sinking (a very handy fact for free divers).

Perhaps in a fresh water. In a sea water at these depths I am positively buoyant even with a 2 kg lead weights on my belt.

 

[Charles123]

Sophiecentaur, thank you for your answer. Two things come to mind, first is the problem with the buoyant force that will not be constant, because the volume of the body will decrease with depth because of compression; second, the way in enter water is important for hydrodynamic resistance, so you have to sum to the buoyant force the hydrodynamic force resistance (1/2*rowater*Velocity^2*Cd*Area, being that velocity will also not be constant, as it will decrease with depth because of resistance and boyant force). Is this not the case? How should one deal with this problems?
Regards